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Surprisingly, we got only one question for our 2-hour exam and I think nobody solved it. Here is the problem:
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Addition: Trivial. Any sum of two finite sums of squares is a finite sum of squares. Multiplication: Trivial. Any product of two squares is a square and the distributive property shows that a product of two finite sums has finitely many terms. Division: Suppose that $a=\sum_{i=1}^n x_i^2$ is not zero. We would like to show that $a^{-1}\in S$ (using s.harp's comment). If $n=1$, then the inverse of $a$ is trivial. The first interesting case is when $n=2$ Suppose that $a=x_1^2+x_2^2$. Now, $$ \frac{1}{x_1^2+x_2^2}=\frac{x_1^2+x_2^2}{(x_1^2+x_2^2)^2}=\left(\frac{x_1}{x_1^2+x_2^2}\right)^2+\left(\frac{x_2}{x_1^2+x_2^2}\right)^2 $$ The denominators are OK because this is a field. All higher $n$'s are generalizations of this case. |
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$$\frac{1}{x_1^2+x_2^2+x_3^2}=\frac{x_1^2}{(x_1^2+x_2^2+x_3^2)^2}+\frac{x_2^2}{(x_1^2+x_2^2+x_3^2)^2}+\frac{x_3^2}{(x_1^2+x_2^2+x_3^2)^2}$$ |
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$\begin{align}{\bf Hint} \ \ a\neq 0\ \Rightarrow&\ \ \color{#c00}{a^{\large -2}} =\, (a^{\large -1})^{\large 2} \in S\ \ \text{by $\,S\,$ contains all squares}\\ {\rm so}\ \ \ a\in S\ \Rightarrow&\ \ a^{\large -1} =\, a\cdot \color{#c00}{a^{\large-2}}\in S\ \ \text{by $\,S\,$ is closed under multipication} \end{align}$ Remark $ $ The essence can be presented more simply in additive form, using only integers. Theorem $\ $ If $\,S\,$ is a set of integers containing all even integers, Proof $\ $ Suppose $\,a,b\in S.\,$ If $\,S\,$ is closed under addition then $\,a+b+2(-b) = a-b\in S\,$ since by hypothesis the even integer $\,2(-b)\in S.\,$ Conversely if $\,S\,$ is closed under subtraction then $\,a-(0-b) = a+b\in S\,$ since by hypothesis the even integer $\,0\in S.$ Readers familiar with groups and monoids can rephrase the above in that language. |
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